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I am stuck with something that may be simple, but I just cannot find the answer.

We define, for $z$ a complex number of real part strictly positive, $\Gamma(z) = \int_{0}^{\infty} t^{z - 1}exp(-t) dt$. How do we show that $\Gamma$ is an analytic function on $\{ z \in \mathbb{C}, \Re(z) > 0 \}$ ?

I started by rewritting the expression of $\Gamma$ this way:

$\Gamma(z) = \int_{- \infty}^{\infty} exp(zu)exp(-e^{u}) du$

Then, if $a$ is a complex number with $\Re(a) > 0$, we have:

$\Gamma(z) = \int_{- \infty}^{\infty} exp((z - a)u)exp(au)exp(-e^{u}) du = \int_{- \infty}^{\infty} \sum \limits_{n = 0}^{+\infty} \frac{(z-a)^{n}}{n!}u^{n}exp(-e^{u}) du$.

However after that, I do not see how I could exchange $\sum$ and $\int$ ...

Thank you for your help.

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1 Answer 1

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It is analytical, because it is differentiable as a function of a complex variable $z$. The integral corresponding to the derivative converges uniformly, so you can differentiate with respect to the parameter.

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  • $\begingroup$ Thank you for your answer. However, I am not actually supposed to know anything about complex analysis (I only have theorems about real analysis, which for some of them work with complex functions as well, but differentiating a complex function is not part of that). Is it not possible to conclude with what I started ? $\endgroup$
    – JackEight
    Feb 4, 2021 at 20:25
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    $\begingroup$ @JackEight Well, if you're trying to prove that $\Gamma$ is just a real analytic function, use the same approach: differentiate the integral w.r.t. the parameter and prove that all obtained integrals for all derivatives converge uniformly. But if you want to prove the statement for $\{z \in \mathbb{C}: \Re (z) > 0\}$, you should know some basic complex analysis. $\endgroup$
    – Timur Bakiev
    Feb 4, 2021 at 20:30

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